I am trying to run while loop in shell
NODESTATE="0"
LOOPC="1"
while [ "$NODESTATE" -ne "UP" ]; do
echo "node is up "
but it is throwing me an error with [: UP: integer expression expected
or shoud i use != instead of -ne
The problem is in your condition:
while [ "$NODESTATE" -ne "UP" ]; do
The -ne option is used to determine whether one comparator is numerically equal to another. But you're doing a string comparison, not a numeric one. Instead, try the following:
while [ "$NODESTATE" != "UP" ]; do
You can read man test to see how the various options to [ work.
Related
I am missing something fundamental concerning either the bash's if construct/operators or string comparison.
Consider the following script:
#!/bin/bash
baseSystem="testdir1"
testme="NA"
if [ "$baseSystem"=="$testme" ]; then
echo "In error case"
fi
if [ "$baseSystem"!="$testme" ]; then
echo "In error case"
fi
I get:
In error case
In error case
So it enters each case even though they should be mututally exclusive.
Any help is appreciated.
bash happens to be somewhat particular about spaces.
Add spaces around the operators:
if [ "$baseSystem" == "$testme" ]; then
...
if [ "$baseSystem" != "$testme" ]; then
The following are not equivalent:
[ "$a"="$b" ]
[ "$a" = "$b" ]
Your first test is essentially the same as saying if [ "testdir1==NA" ]; then which would always be true.
I couldn't find any one simple straightforward resource spelling out the meaning of and fix for the following BASH shell error, so I'm posting what I found after researching it.
The error:
-bash: [: too many arguments
Google-friendly version: bash open square bracket colon too many arguments.
Context: an if condition in single square brackets with a simple comparison operator like equals, greater than etc, for example:
VARIABLE=$(/some/command);
if [ $VARIABLE == 0 ]; then
# some action
fi
If your $VARIABLE is a string containing spaces or other special characters, and single square brackets are used (which is a shortcut for the test command), then the string may be split out into multiple words. Each of these is treated as a separate argument.
So that one variable is split out into many arguments:
VARIABLE=$(/some/command);
# returns "hello world"
if [ $VARIABLE == 0 ]; then
# fails as if you wrote:
# if [ hello world == 0 ]
fi
The same will be true for any function call that puts down a string containing spaces or other special characters.
Easy fix
Wrap the variable output in double quotes, forcing it to stay as one string (therefore one argument). For example,
VARIABLE=$(/some/command);
if [ "$VARIABLE" == 0 ]; then
# some action
fi
Simple as that. But skip to "Also beware..." below if you also can't guarantee your variable won't be an empty string, or a string that contains nothing but whitespace.
Or, an alternate fix is to use double square brackets (which is a shortcut for the new test command).
This exists only in bash (and apparently korn and zsh) however, and so may not be compatible with default shells called by /bin/sh etc.
This means on some systems, it might work from the console but not when called elsewhere, like from cron, depending on how everything is configured.
It would look like this:
VARIABLE=$(/some/command);
if [[ $VARIABLE == 0 ]]; then
# some action
fi
If your command contains double square brackets like this and you get errors in logs but it works from the console, try swapping out the [[ for an alternative suggested here, or, ensure that whatever runs your script uses a shell that supports [[ aka new test.
Also beware of the [: unary operator expected error
If you're seeing the "too many arguments" error, chances are you're getting a string from a function with unpredictable output. If it's also possible to get an empty string (or all whitespace string), this would be treated as zero arguments even with the above "quick fix", and would fail with [: unary operator expected
It's the same 'gotcha' if you're used to other languages - you don't expect the contents of a variable to be effectively printed into the code like this before it is evaluated.
Here's an example that prevents both the [: too many arguments and the [: unary operator expected errors: replacing the output with a default value if it is empty (in this example, 0), with double quotes wrapped around the whole thing:
VARIABLE=$(/some/command);
if [ "${VARIABLE:-0}" == 0 ]; then
# some action
fi
(here, the action will happen if $VARIABLE is 0, or empty. Naturally, you should change the 0 (the default value) to a different default value if different behaviour is wanted)
Final note: Since [ is a shortcut for test, all the above is also true for the error test: too many arguments (and also test: unary operator expected)
Just bumped into this post, by getting the same error, trying to test if two variables are both empty (or non-empty). That turns out to be a compound comparison - 7.3. Other Comparison Operators - Advanced Bash-Scripting Guide; and I thought I should note the following:
I used -e thinking it means "empty" at first; but that means "file exists" - use -z for testing empty variable (string)
String variables need to be quoted
For compound logical AND comparison, either:
use two tests and && them: [ ... ] && [ ... ]
or use the -a operator in a single test: [ ... -a ... ]
Here is a working command (searching through all txt files in a directory, and dumping those that grep finds contain both of two words):
find /usr/share/doc -name '*.txt' | while read file; do \
a1=$(grep -H "description" $file); \
a2=$(grep -H "changes" $file); \
[ ! -z "$a1" -a ! -z "$a2" ] && echo -e "$a1 \n $a2" ; \
done
Edit 12 Aug 2013: related problem note:
Note that when checking string equality with classic test (single square bracket [), you MUST have a space between the "is equal" operator, which in this case is a single "equals" = sign (although two equals' signs == seem to be accepted as equality operator too). Thus, this fails (silently):
$ if [ "1"=="" ] ; then echo A; else echo B; fi
A
$ if [ "1"="" ] ; then echo A; else echo B; fi
A
$ if [ "1"="" ] && [ "1"="1" ] ; then echo A; else echo B; fi
A
$ if [ "1"=="" ] && [ "1"=="1" ] ; then echo A; else echo B; fi
A
... but add the space - and all looks good:
$ if [ "1" = "" ] ; then echo A; else echo B; fi
B
$ if [ "1" == "" ] ; then echo A; else echo B; fi
B
$ if [ "1" = "" -a "1" = "1" ] ; then echo A; else echo B; fi
B
$ if [ "1" == "" -a "1" == "1" ] ; then echo A; else echo B; fi
B
Another scenario that you can get the [: too many arguments or [: a: binary operator expected errors is if you try to test for all arguments "$#"
if [ -z "$#" ]
then
echo "Argument required."
fi
It works correctly if you call foo.sh or foo.sh arg1. But if you pass multiple args like foo.sh arg1 arg2, you will get errors. This is because it's being expanded to [ -z arg1 arg2 ], which is not a valid syntax.
The correct way to check for existence of arguments is [ "$#" -eq 0 ]. ($# is the number of arguments).
I also faced same problem. #sdaau answer helped me in logical way. Here what I was doing which seems syntactically correct to me but getting too many arguments error.
Wrong Syntax:
if [ $Name != '' ] && [ $age != '' ] && [ $sex != '' ] && [ $birthyear != '' ] && [ $gender != '' ]
then
echo "$Name"
echo "$age"
echo "$sex"
echo "$birthyear"
echo "$gender"
else
echo "Enter all the values"
fi
in above if statement, if I pass the values of variable as mentioned below then also I was getting syntax error
export "Name"="John"
export "age"="31"
export "birthyear"="1990"
export "gender"="M"
With below syntax I am getting expected output.
Correct syntax:
if [ "$Name" != "" -a "$age" != "" -a "$sex" != "" -a "$birthyear" != "" -a "$gender" != "" ]
then
echo "$Name"
echo "$age"
echo "$sex"
echo "$birthyear"
echo "$gender"
else
echo "it failed"
fi
There are few points which we need to keep in mind
use "" instead of ''
use -a instead of &&
put space before and after operator sign like [ a = b], don't use as [ a=b ] in if condition
Hence above solution worked for me !!!
Some times If you touch the keyboard accidentally and removed a space.
if [ "$myvar" = "something"]; then
do something
fi
Will trigger this error message. Note the space before ']' is required.
I have had same problem with my scripts. But when I did some modifications it worked for me. I did like this :-
export k=$(date "+%k");
if [ $k -ge 16 ]
then exit 0;
else
echo "good job for nothing";
fi;
that way I resolved my problem. Hope that will help for you too.
while [ $done = 0 ]
do
echo -n "Would you like to create one? [y/n]: "
read answer
if [ "$(answer)" == "y" ] || [ "$(answer)" == "Y" ]; then
mkdir ./fsm_$newVersion/trace
echo "Created trace folder in build $newVersion"
$done=1
elif [ "$(answer)" == "n" ] || [ "$(answer)" == "N" ]; then
$done=2
else
echo "Not a valid answer"
fi
done
Ok so I have this simple bashscript above that simply just tries to get input from a user and validate it. However I keep getting this error
./test.sh: line 1: answer: command not found
./test.sh: line 1: answer: command not found
./test.sh: line 1: answer: command not found
./test.sh: line 1: answer: command not found
Which I have no idea why because "answer" is nowhere near line 1. So I ran into this article
Which makes sense since it's referring to line 1 and can't find answer. So it seems to be starting a new subshell. However I didn't really understand the solution and can't see how I would apply it to my case. I just wanna get this to work.
$(answer) doesn't substitute the value of the variable answer. It executes answer as a command, and substitutes the output of that command. You want ${answer} everywhere you have $(answer). In this case you can get away with bare $answer too, but overuse of ${...} is good paranoia.
(Are you perhaps used to writing Makefiles? $(...) and ${...} are the same in Makefiles, but the shell is different.)
By the way, you have some other bugs:
In shell, you do not put a dollar sign on the variable name on the left hand side of an assignment. You need to change $done=1 to just done=1 and similarly for $done=2.
You are not being paranoid enough about your variable substitutions. Unless you know for a fact that it does the wrong thing in some specific case, you should always wrap all variable substitutions in double quotes. This affects both the mkdir command and the condition on the while loop.
You are not being paranoid enough about arguments to test (aka [). You need to prefix both sides of an equality test with x so that they cannot be misinterpreted as switches.
== is not portable shell, use = instead (there is no difference in bash, but many non-bash shells do not support == at all).
Put it all together and this is what your script should look like:
while [ "x${done}" = x0 ]; do
echo -n "Would you like to create one? [y/n]: "
read answer
if [ "x${answer}" = xy ] || [ "x${answer}" = xY ]; then
mkdir "./fsm_${newVersion}/trace"
echo "Created trace folder in build $newVersion"
done=1
elif [ "x${answer}" = xn ] || [ "x${answer}" = xN ]; then
done=2
else
echo "Not a valid answer"
fi
done
Which I have no idea why because
"answer" is nowhere near line 1. So I
ran into this article
That's not your problem here.
I ran the script and did not get the error you got. I did receive the error:
./test.sh: line 1: [: -eq: unary operator expected
when I tried to compile though. Defining done fixed this. The following script should work...
#!/bin/bash
done=0
while [ $done -eq 0 ]
do
echo -n "Would you like to create one? [y/n]: "
read answer
if [[ "$(answer)" == "y" || "$(answer)" == "Y" ]]; then
mkdir ./fsm_${newVersion}/trace
echo "Created trace folder in build $newVersion"
$done=1
elif [[ "$(answer)" == "n" || "$(answer)" == "N" ]]; then
$done=2
else
echo "Not a valid answer"
fi
done
...note you were doing string comparisons on your done variable, which you apparently intended to be numeric. It's generally bad form to do string comparison on a numeric type variable, though it will work. Use -eq (arithmetic comparison operator) instead. (Also note that if you kept that test, your string equality would be inconsistent... you had "=" in one spot and "==" in another spot... nitpicking here, but it's helpful to be consistent).
Also, I suggest double brackets for your compound conditionals as they will be more readable if you have longer ones. e.g.
if [[($var1 -eq 0 && $var2 -eq 1) || ($var1 -eq 1 && $var2 -eq 0)]]; then
Just a matter of preference as you only have two conditions, but could be useful in the future.
Also you were missing braces '{' '}' around your newVersion variable.
Finally, I'd suggest putting the line #!/bin/bash on the top of your script. Otherwise its up to your environment to determine what to do with your script, which is a bad idea.
Usually work in Windows, but trying to setup RabbitMQ on my Mac. Can someone let me know what the line below does?
[ "x" = "x$RABBITMQ_NODE_IP_ADDRESS" ] && [ "x" != "x$NODE_IP_ADDRESS" ] && RABBITMQ_NODE_IP_ADDRESS=${NODE_IP_ADDRESS}
Specifically, I'm curious about the [ "x" = "x$RAB..."] syntax.
If the RABBITMQ_NODE_IP_ADDRESS variable is empty/doesn't exist, it'll evaluate as "x" = "x" , which is true.
So it basically says, if RABBITMQ_NODE_IP_ADDRESS isn't set and NODE_IP_ADDRESS is set, set RABBITMQ_NODE_IP_ADDRESS=NODE_IP_ADDRESS
The "x" is used (somewhat superstitiously*) to prevent errors if the variable is null or unset. Most of the time the quotes take care of that for you. By putting the literal first and the variable second you eliminate errors in cases where the variable contains a string that starts with a dash, since test (aka [) would think it is an operator. In the case of your example, it would be preferable to use the -z and -n operators that test whether a variable is empty (null or unset) or not empty, respectively.
POSIX shells, such as Bourne (works in Bash, too):
[ -z $RABBITMQ_NODE_IP_ADDRESS ] && [ -n $NODE_IP_ADDRESS" ] && RABBITMQ_NODE_IP_ADDRESS=${NODE_IP_ADDRESS}
Bash (and ksh and zsh):
[[ -z $RABBITMQ_NODE_IP_ADDRESS && -n $NODE_IP_ADDRESS" ]] && RABBITMQ_NODE_IP_ADDRESS=${NODE_IP_ADDRESS}
* There may be some shells that need the "x", but some people do that "because it's always been done that way".
The "x" is not always superstitious, even in my relatively new bash (4.0.33).
Let's put the operation between parens. Empty variables are fine:
$ a=""
$ b=""
$ if [ '(' "$a" = "$b" ')' ]; then echo both_equal; fi
both_equal
But the ! operator for instance is not:
$ a='!'
$ if [ '(' "$a" = "$b" ')' ]; then echo both_equal; fi
bash: [: `)' expected, found
This is not a problem if we write "x$a" = "x$b" instead of "$a" = "$b".
The bracket [ is a test operator, which you can think of as an if statement. This is checking to see if the shell variable RABBITMQ_NODE_IP_ADDRESS is empty. Unfortunately, if you try to compare to an empty string "", the shell eliminates it before it does the test and your binary comparison operator only gets one (or maybe zero) operands. To prevent that error, it is a common practice to concatenate an "x" on each side of the =. Thus, instead of
[ "" = "<variable>" ]
becoming
[ = value ]
and yielding an error,
[ "X" = "X<variable>" ]
becomes
[ X = Xvalue ]
and the comparison may continue
i am testing with the shell script below:
#!/bin/ksh -x
instance=`echo $1 | cut -d= -f2`
if [ $instance == "ALL" ]
then
echo "strings matched \n"
fi
It's giving this error in the if condition:
: ==: unknown test operator
is == really not the correct syntax to use?
I am running on the command line as below
test_lsn_2 INSTANCE=ALL
Could anybody please suggest a solution.
Thanks.
To compare strings you need a single =, not a double. And you should put it in double quotes in case the string is empty:
if [ "$instance" = "ALL" ]
then
echo "strings matched \n"
fi
I see that you are using ksh, but you added bash as a tag, do you accept a bash-related answer?
Using bash you can do it in these ways:
if [[ "$instance" == "ALL" ]]
if [ "$instance" = "ALL" ]
if [[ "$instance" -eq "ALL" ]]
See here for more on that.
Try
if [ "$instance" = "ALL" ]; then
There were several mistakes:
You need double quotes around the variable to protect against the (unlikely) case that it's empty. In this case, the shell would see if [ = "ALL" ]; then which isn't valid.
Equals in the shell uses a single = (there is no way to assign a value in an if in the shell).
totest=$1
case "$totest" in
"ALL" ) echo "ok" ;;
* ) echo "not ok" ;;
esac
I'va already answered a similar question. Basically the operator you need is = (not ==) and the syntax breaks if your variable is empty (i.e. it becomes if [ = ALL]). Have a look at the other answer for details.